Heat and Specific Heat
Heat transfer, specific heat capacity, and calorimetry
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🔥 Heat and Specific Heat
Heat vs. Temperature
Temperature: Measure of average kinetic energy of particles Heat (Q): Energy transferred between objects due to temperature difference
💡 Key Distinction: Temperature is a property; heat is energy in transit.
Heat flows spontaneously from hot to cold until thermal equilibrium is reached.
Units of Heat
- Joule (J): SI unit of energy
- Calorie (cal): Energy to raise 1 g of water by 1°C
- 1 cal = 4.186 J
- Kilocalorie (kcal or Cal): Food "Calorie"
- 1 kcal = 1000 cal = 4186 J
Specific Heat Capacity
Specific heat (c) is the energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).
where:
- = heat transferred (J)
- = mass (kg)
- = specific heat (J/kg·°C or J/kg·K)
- = temperature change (°C or K)
Common Specific Heats:
| Substance | c (J/kg·°C) | |-----------|-------------| | Water | 4186 | | Ice | 2090 | | Steam | 2010 | | Aluminum | 900 | | Steel | 470 | | Copper | 387 | | Lead | 128 |
Water has the highest specific heat of common substances! This is why:
- Coastal areas have moderate climates
- Water is used in cooling systems
- Land heats/cools faster than oceans
Sign Convention
- : Heat absorbed (temperature increases)
- : Heat released (temperature decreases)
If : temperature increases → heat absorbed If : temperature decreases → heat released
Calorimetry
Calorimetry measures heat transfer using conservation of energy.
Principle: In an isolated system, heat lost = heat gained
or
Typical Setup:
- Hot object placed in cold water
- System isolated (insulated calorimeter)
- Final temperature measured
- Energy conservation:
Heat Capacity
Heat capacity (C) is heat needed to raise an object's temperature by 1°C:
Units: J/°C
Difference:
- Specific heat (c): Property of material (per kg)
- Heat capacity (C): Property of specific object (total)
Methods of Heat Transfer
1. Conduction
- Heat transfer through direct contact
- Molecular collisions
- Requires material medium
- Example: Metal spoon in hot soup
2. Convection
- Heat transfer by fluid motion
- Hot fluid rises, cold sinks
- Requires fluid (liquid or gas)
- Example: Boiling water, ocean currents
3. Radiation
- Heat transfer by electromagnetic waves
- No medium required (works in vacuum)
- All objects emit thermal radiation
- Example: Sun warming Earth, heat lamps
Problem-Solving Strategy
- Identify all objects exchanging heat
- Set up energy conservation:
- Write Q for each object:
- Determine signs:
- Heating: ,
- Cooling: ,
- Solve for unknown (usually final temperature or specific heat)
- Check reasonableness: Final T should be between initial temperatures
Common Mistakes
❌ Using wrong sign for heat (lost vs. gained) ❌ Forgetting to convert units (g → kg, cal → J) ❌ Using Celsius in place of Kelvin inappropriately (ΔT is same, but not absolute T) ❌ Neglecting heat lost to surroundings (real calorimeters aren't perfectly insulated) ❌ Mixing up specific heat (c) and heat capacity (C)
📚 Practice Problems
1Problem 1easy
❓ Question:
How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? (c_water = 4186 J/kg·°C)
💡 Show Solution
Given:
- Mass: kg
- Initial temp:
- Final temp:
- Specific heat: J/kg·°C
Find: Heat required
Solution:
Step 1: Calculate temperature change.
Step 2: Apply heat formula.
Answer: 502 kJ of heat is required
This is equivalent to about 120 food Calories (kcal).
2Problem 2medium
❓ Question:
A 0.50 kg piece of aluminum at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature? (c_Al = 900 J/kg·°C, c_water = 4186 J/kg·°C)
💡 Show Solution
Given:
- Aluminum: kg, , J/kg·°C
- Water: kg, , J/kg·°C
Find: Final temperature
Solution:
Step 1: Set up energy conservation.
Step 2: Express temperature changes.
Step 3: Substitute values.
Verification:
- Aluminum cools: ✓
- Water warms: ✓
- Final T between initial temperatures ✓
Answer: Final temperature is 24.1°C
The water barely warms because it has much larger mass and specific heat!
3Problem 3hard
❓ Question:
A 0.20 kg piece of unknown metal at 150°C is placed in 0.50 kg of water at 20°C. The final temperature is 25°C. What is the specific heat of the metal? Assume no heat is lost to surroundings. (c_water = 4186 J/kg·°C)
💡 Show Solution
Given:
- Metal: kg, ,
- Water: kg, , J/kg·°C
- Final:
Find: Specific heat of metal
Solution:
Step 1: Set up energy conservation.
Step 2: Calculate temperature changes.
Step 3: Solve for .
Comparison with known metals:
- Copper: 387 J/kg·°C
- Steel: 470 J/kg·°C
Answer: Specific heat is 419 J/kg·°C
This is close to copper (387) or possibly a copper alloy.